9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Chapter 4.3, Problem 41AYU

To determine

To analyze: the graph of given function

Expert Solution & Answer

Answer to Problem 41AYU

Precalculus, Chapter 4.3, Problem 41AYU , additional homework tip 1

Explanation of Solution

Given:

$f(x)=x2+1x$

Calculation:

STEP 1: Rewrite the function as $f(x)=x3+1x$

Now, factor the function $f$ .

$f(x)=(x+1)(x2−x+1)x$

The domain of $f$ is ${x|x=0}$ .

Since $0$ is in the domain, there is no $y−intercept$ .

STEP 2: The function $f$ is in lowest terms. Since $x+1=0$ has the real solution $x=−1,−1 is the x−intercept$ .

Determine the behavior of the graph near the $x−intercept$ .

$R(x)=x3+1−1≈−1(x3+1)$

Plot the point $(−1, 0)$ and indicate a line with slope $−1$ on the graph.

STEP 3: $f$ is in lowest terms. So, $x=0$ is a vertical asymptote of $f$ . Graph the line $x=0$ using a dashed line.

STEP 4: Degree of the numerator is greater than the degree of the denominator. So, the rational function is improper. To find any horizontal or oblique asymptote, we use the long division.

$xx2x3+1x3 1$

The quotient is $x2$ . So, the graph does not have any horizontal or oblique asymptote. But the graph of $f$ will approach the graph of $y=x2 as x→−∞ and x→∞$ . The graph of $f$ does not intersect with $y=x2$ .

STEP 5: The numerator has one zero at $−1$ and the denominator has one zero at $0$ . Therefore, we divide the $x−axis$ into three intervals. The three intervals are $(−∞, −1), (−1, 0), (0, ∞)$ .

Now, construct the table.

Precalculus, Chapter 4.3, Problem 41AYU , additional homework tip 2

Interval Number chosen Value of f Location of graph Point on graph

$(−∞,−1)−2f(−2)=3.5Above x−axis(−2,3.5)$

$(−1,0)−12 f(−12)=−1.75Below x−axis(−12,−1.75)$

$(0,∞)1f(1)=2Above x−axis(1,2)$

Plot the points from the table obtained.

Precalculus, Chapter 4.3, Problem 41AYU , additional homework tip 3

STEP 6: Since the graph of $f$ is above the $x−axis for x<−1$ , we place arrows above the $x−axis$ as shown in the figure. Similarly, as the graph of $f$ is below the $x−axis for −1<x<0$ , we place arrows below the $x−axis$ . Also, as the graph of $f$ is above the $x−axis for x>0$ , we place arrows above the $x−axis$ as shown in the figure.

Precalculus, Chapter 4.3, Problem 41AYU , additional homework tip 4

STEP 7: The figure shows the final graph.

Precalculus, Chapter 4.3, Problem 41AYU , additional homework tip 5

Conclusion:

Thus, the graph of $f$ is obtained which illustrates the information obtained from Steps 1 through 6.

Chapter 4.3, Problem 40AYU

Chapter 4.3, Problem 42AYU

Chapter 4 Solutions

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